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简单线段树操作

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Transformation HDU - 4578 vj

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#include <iostream>
#include <cstring>
using namespace std;
typedef long long ll;
const ll N = 2e5+99;
const ll mod = 10007;



struct segment_tree {
    #define pl (p << 1)
    #define pr (p << 1 | 1)
    struct tree {
       ll data, data2 ,data3; 
       ll add;
       ll c;
       ll mul;
    }tr[N << 2];
    inline void pushup(ll p) {
        (tr[p].data = tr[pl].data + tr[pr].data)%=mod;
        (tr[p].data2 = tr[pl].data2 + tr[pr].data2) %= mod;
        (tr[p].data3 = tr[pl].data3 + tr[pr].data3) %= mod;
    }
    inline void pushdown(ll s, ll t, ll p) {
        ll mid = (s + t) >> 1;
        if (tr[p].c != -1) {
            (tr[pl].data = tr[p].c * (mid -s + 1))%=mod;
            (tr[pl].data2 = tr[p].c*tr[p].c%mod * (mid - s + 1)) %= mod;
            (tr[pl].data3 = tr[p].c *tr[p].c%mod*tr[p].c % mod * (mid - s + 1)) %= mod;
            (tr[pr].data = tr[p].c * (t - mid)) %= mod;
            (tr[pr].data2 = tr[p].c * tr[p].c % mod * (t - mid)) %= mod;
            (tr[pr].data3 = tr[p].c *tr[p].c%mod*tr[p].c % mod * (t - mid)) %= mod;
            tr[pl].c  = tr[p].c;
            tr[pr].c  = tr[p].c;
            if (tr[p].c != -1) {
                tr[pl].add = tr[pr].add = 0;
                tr[pl].mul = tr[pr].mul = 1;
            }
        }
        (tr[pl].data *= tr[p].mul)%=mod;
        (tr[pl].data2 *= tr[p].mul*tr[p].mul%mod) %= mod;
        (tr[pl].data3 *= tr[p].mul*tr[p].mul%mod * tr[p].mul%mod) %= mod;
        (tr[pr].data *= tr[p].mul) %= mod;
        (tr[pr].data2 *= tr[p].mul * tr[p].mul % mod) %= mod;
        (tr[pr].data3 *= tr[p].mul * tr[p].mul % mod * tr[p].mul % mod) %= mod;
        (tr[pr].mul *= tr[p].mul)%=mod;
        (tr[pl].mul *= tr[p].mul)%=mod;
        

        (tr[pl].data3 += 3*tr[p].add * tr[pl].data2%mod+3*tr[p].add*tr[p].add%mod*tr[pl].data%mod + tr[p].add*tr[p].add%mod*tr[p].add%mod*(mid - s+1)%mod)%=mod;
        (tr[pl].data2 += 2 * tr[p].add*(tr[pl].data)%mod+tr[p].add*tr[p].add%mod*(mid - s+1))%=mod;
        (tr[pl].data += tr[p].add * (mid - s+1))%=mod;


        (tr[pr].data3 += 3*tr[p].add * tr[pr].data2%mod+3*tr[p].add*tr[p].add%mod*tr[pr].data%mod + tr[p].add*tr[p].add%mod*tr[p].add%mod*(t - mid)%mod)%=mod;
        (tr[pr].data2 += 2 * tr[p].add*(tr[pr].data)%mod+tr[p].add*tr[p].add%mod*(t - mid))%=mod;
        (tr[pr].data += tr[p].add * (t - mid))%=mod;


        (tr[pl].add *= tr[p].mul)%=mod;
        (tr[pr].add *= tr[p].mul)%=mod;
        (tr[pl].add += tr[p].add)%=mod;
        (tr[pr].add += tr[p].add)%=mod;


        tr[p].mul = 1;
        tr[p].add = 0;
        tr[p].c = -1;
    }
    inline void add(ll s, ll t, ll l, ll r, ll p, ll c) {
        if (s >= l &&t <= r) {
            (tr[p].add+=c)%=mod;
            (tr[p].data3 += 3*c*tr[p].data2%mod+3*tr[p].data*c%mod*c%mod+c*c%mod*c%mod* ( t - s + 1)%mod) %= mod;
            (tr[p].data2 += 2*c%mod * tr[p].data % mod + c * c % mod * (t - s + 1)%mod) %= mod;
            (tr[p].data += c * ( t - s + 1)) %= mod;
            return;
        }        
        pushdown(s, t, p);
        ll mid = (s + t) >> 1;
        if (l <= mid)add(s, mid, l, r, pl, c);
        if (r >= mid + 1)add(mid + 1, t, l, r, pr, c);
        pushup(p);
    }
    inline void mul(ll s, ll t, ll l, ll r, ll p, ll c) {
        if (s >= l && t <= r) {
            (tr[p].mul *= c)%=mod;
            (tr[p].add *= c)%=mod;
            (tr[p].data *= c)%=mod;
            (tr[p].data2 *= c*c%mod)%=mod;
            (tr[p].data3 *= c*c%mod*c%mod)%=mod;
            return;
        } 
        pushdown(s, t, p);
        ll mid = (s + t) >> 1;
        if (l <= mid)mul(s, mid, l, r, pl, c);
        if (r >= mid + 1)mul(mid + 1, t, l, r,pr, c );
        pushup(p);
    }
    inline void change(ll s, ll t, ll l, ll r, ll p, ll c) {
        if (s >= l && t <= r) {
            tr[p].c = c;
            tr[p].add = 0;
            tr[p].mul = 1;
            (tr[p].data = c*(t-s+1))%=mod;
            (tr[p].data2 = c*c%mod*(t-s+1))%=mod;
            (tr[p].data3 = c*c%mod*c%mod*(t-s+1)%mod)%=mod;
            return;
        }
        pushdown(s, t, p);
        ll mid = (s + t ) >> 1;
        if (l <= mid)change(s, mid, l ,r, pl, c);
        if (r > mid)change(mid + 1, t, l, r, pr, c);
        pushup(p);
    }
    inline ll query(ll s, ll t, ll l, ll r, ll p, ll c) {
        ll mid = (s + t >> 1);
        pushdown(s, t, p);
        if (s >= l && t <= r)  {
            if (c == 1)
            return tr[p].data;
            if(c == 2)return tr[p].data2;
            if (c == 3)return tr[p].data3;
        }
        ll ret = 0;
        if (l <= mid) (ret += query(s, mid, l, r, pl, c))%=mod;
        if (r > mid) (ret += query(mid + 1, t, l, r, pr, c))%=mod;
        return ret;
       
    }
    inline void build(ll s, ll t, ll p) {
        tr[p].add = tr[p].data = tr[p].data2 = tr[p].data3 = 0;
        tr[p].c = -1;
        tr[p].mul = 1;
        if (s == t) {
            tr[p].data = 0;
            return;
        }
        ll mid = s+ t>>1;
        build(s, mid, pl);
        build(mid + 1, t, pr);
        pushup(p);
        return;
    }
}T;

int main() {
    ll n, m;
    ios::sync_with_stdio(0);
    while (cin >> n >> m) {
        if (n == 0 && m == 0)break;
        memset(T.tr, 0, sizeof T.tr);
        T.build(1, n, 1);
        while (m--) {
            ll opt;cin >> opt;
            ll x, y, c;
            cin >> x >> y >> c;
            if (opt == 1) 
                T.add(1, n, x, y, 1, c%mod);
            if (opt == 2) 
                T.mul(1, n, x, y, 1, c%mod);
            if (opt == 3)
                T.change(1, n, x, y, 1, c%mod);
            if (opt == 4)
                cout << T.query(1, n, x, y, 1, c)  <<‘\n‘;
        }
    }
    return 0;
}

Transformation HDU - 4578

标签：orm   tran   cout   操作   href   str   iostream   while   problem   

原文地址：https://www.cnblogs.com/Xiao-yan/p/14157025.html